8.1: a) s.mkg1020.1)smkg)(12.0000,10( 5  b) (i) Five times the speed, s.m0.60 (ii) 5   s.m8.26sm0.12  8.2: See Exercise 8.3 (a); the iceboats have the same kinetic energy, so the boat with the larger mass has the larger magnitude of momentum by a factor of .2)()2( mm 8.3: a) . 2 1 2 1 2 1 222 2 m p m vm mvK  b) From the result of part (a), for the same kinetic energy, 2 2 2 1 2 1 m p m p  , so the larger mass baseball has the greater momentum;   .525.0145.0040.0 ballbird pp From the result of part (b), for the same momentum 2211 mKmK  , so 2211 wKwK  ; the woman, with the smaller weight, has the larger kinetic energy.   .643.0700450 womanman KK 8.4: From Eq. (8.2),    smkg1.7820.0cossm50.4kg420.0  xx mvp    s.mkg0.64620.0sin sm50.4kg420.0  yy mvp 8.5: The y-component of the total momentum is       s.mkg256.0sm80.7kg0570.0sm30.1kg145.0  This quantity is negative, so the total momentum of the system is in the y -direction. 8.6: From Eq. (8.2),    s,mkg015.1sm00.7kg145.0  y p and    s,mkg405.0sm00.9kg045.0  x p so the total momentum has magnitude     s,mkg09.1smkg015.1smkg405.0 22 22  yx ppp and is at an angle arctan      68 405. 015.1 , using the value of the arctangent function in the fourth quadrant   .0p,0 y  x p 8.7:    N.563 s1000.2 sm0.25kg0450.0 3      t p The weight of the ball is less than half a newton, so the weight is not significant while the ball and club are in contact. 8.8: a) The magnitude of the velocity has changed by     s,m0.100sm0.55sm0.45  and so the magnitude of the change of momentum is s,mkg14.500)sm0.100(kg)145.0(  to three figures. This is also the magnitude of the impulse. b) From Eq. (8.8), the magnitude of the average applied force is s1000.2 kg.m/s500.14 3  =7.25 N.10 3  8.9: a) Considering the +x-components, ,smkg73.1s)05.0(N)0.25(s)m00.3kg)(16.0( 12  Jpp and the velocity is 10.8 sm in the +x-direction. b) p 2 = 0.48 smkg  + (–12.0 N)(0.05 s) = –0.12 smkg  , and the velocity is +0.75 sm in the –x-direction. 8.10: a) F  t=(1.04 j  )smkg10 5  . b) (1.04 j ˆ )smkg10 5  . c) .jj ˆ )sm10.1( ˆ kg)000,95( )kg.1004.1( s m 5   d) The initial velocity of the shuttle is not known; the change in the square of the speed is not the square of the change of the speed. 8.11: a) With ,0 1 t ,)sN1000.2()sN1080.0( 3 2 292 2 7 0 2 ttdtFJ t xx   which is 18.8 smkg  , and so the impulse delivered between t=0 and .i ˆ )smkg(18.8iss1050.2 3 2   t b) ands),1050.2()sm(9.80kg)145.0( 3 2   y J the impulse is j ˆ )smkg1055.3( 3   c) 2 t J x =7.52 N,10 3  so the average force is (7.52 .i ˆ N)10 3  d) jpp    12 ) ˆ 1055.3 ˆ (18.8m/s) ˆ 0.5 ˆ kg)(40.0145.0( 3 jiji   .ji ˆ kg.m/s)73.0( ˆ kg.m/s)0.13(  The velocity is the momentum divided by the mass, or (89.7 m/s) .ji ˆ m/s)0.5( ˆ  8.12: The change in the ball’s momentum in the x-direction (taken to be positive to the right) is m/s,kg15.41m/s)0.5030coss)m0.65((kg)145.0(   so the x- component of the average force is N,108.81 s1075.1 m/skg41.15 3 3     and the y-component of the force is N.107.2 s)1075.1( 30sin m/s)kg)(65.0145.0( 3 3     8.13: a)   2 1 ),( 3 )( 3 1 3 212 t t tt B ttAFdtJ or .0 tif)3/( 1 3 22  tBAtJ b) . 3 3 22 t m B t m A m J m p v  8.14: The impluse imparted to the player is opposite in direction but of the same magnitude as that imparted to the puck, so the player’s speed is cm/s,27.4 kg)0.75( )sm(20.0kg)16.0(  in the direction opposite to the puck’s. 8.15: a) You and the snowball now share the momentum of the snowball when thrown so your speed is s.cm68.5 kg)0.400kg(70.0 s)m(10.0kg)400.0(   b) The change in the snowball’s momentum is s),mkg20.7)sm0.18()kg400.0(  so your speed is cm/s.3.10 kg70.0 smkg20.7   8.16: a) The final momentum is s,mkg1975.0)sm650.0)(350.0()sm120.0)(kg250.0(  taking positive directions to the right. a) Before the collision, puck B was at rest, so all of the momentum is due to puck A’s motion, and m/s.790.0 kg250.0 m/skg5197.0 1    A A m p v b) 2 1 2 2 2 212 2 1 2 1 2 1 AABBAA vmvmvmKKK  2 22 )sm7900.0)(kg250.0( 2 1 )sm650.0)(kg350.0( 2 1 )sm0.120()kg250.0( 2 1   J0023.0 8.17: The change in velocity is the negative of the change in Gretzky’s momentum, divided by the defender’s mass, or s.m66.4 )sm0.13sm50.1( N900 N756 sm00.5 )( 1212    AA B A BB vv m m vv Positive velocities are in Gretzky’s original direction of motion, so the defender has changed direction. b) )( 2 1 )( 2 1 2 1 2 2 2 1 2 212 BBBAAA vvmvvmKK             )m/s)5.00(m/s)N)((4.66090( )m/s)0.13(m/s)N)((1.50675( )m/s08.9(2 1 22 22 2 kJ.85.6 8.18: Take the direction of the bullet’s motion to be the positive direction. The total momentum of the bullet, rifle, and gas must be zero, so ,0m/s)1.85kg)((2.80m/s)1.85m/skg)(60100072.0( gas  p and gas p = 0.866 .smkg  Note that the speed of the bullet is found by subtracting the speed of the rifle from the speed of the bullet relative to the rifle. 8.19: a) See Exercise 8.21;   .sm3.60)sm080.0( kg00.1 kg00.3  A v b) J.46.8m/s)kg)(1.20000.3)(2/1(m/s)(3.60kg)(1.002)1( 22  8.20: In the absence of friction, the horizontal component of the hat-plus-adversary system is conserved, and the recoil speed is .sm66.0 kg)012( 36.9cos)smkg)(22.005.4(   8.21: a) Taking A v and B v to be magnitudes, conservation of momentum is expressed as BBAA vmvm  , so . A B A B v m m v  b) . ))/(()2/1( )2/1( 2 2 2 2 A B ABAB AA BB AA B A m m vmmm vm vm vm K K  (This result may be obtained using the result of Exercise 8.3.) 8.22: XPo: decay Po 2104214214  m/s101.92 kg1065.6 J)102(1.23 2 2 1 : 7 27 12 2             m KE v vmKESetv Momentum conservation: m/s1065.3 kg)1067.1)(210( m/s)10kg)(1.921065.6( 210 0 5 27 727 p xx           m vm m vm v vmvm x x 8.23: Let the +x-direction be horizontal, along the direction the rock is thrown. There is no net horizontal force, so x P is constant. Let object A be you and object B be the rock.  35.0cos0 BBAA vmvm m/s11.2 0.35cos    A BB A m vm v 8.24: Let Rebecca’s original direction of motion be the x-direction. a) From conservation of the x-component of momentum, ,kg)0.65(53.1m/s)coskg)(8.0(45.0m/s)kg)(13.00.45( x v So s.m67.5 x v If Rebecca’s final motion is taken to have a positive -y component, then s.m43.4 kg)0.65( 53.1sin s)mkg)(8.00.45(    y v Daniel’s final speed is  22 yx vv s,m20.7s)m43.4(s)m67.5( 22  and his direction is arctan     38 67.5 42.4 from the -x axis, which is 1.91 from the direction of Rebecca’s final motion. b) 222 s)m(13.0(45.0) 2 1 )sm(7.195kg)0.65( 2 1 s)m(8.0kg)0.45( 2 1 K J.680 Note that an extra figure was kept in the intermediate calculation. 8.25: s),m25.2()sm00.4()sm00.3)(( KenKimKenKim mmmm  so ,750.0 s)m00.3()sm00.4( s)m25.2(s)m00.3( Ken Kim     m m and Kim weighs N.525N)700)(750.0(  8.26: The original momentum is s,mkg1060.9)sm00.4)(kg000,24( 4  the final mass is kg,000,27kg3000kg000,24  and so the final speed is s.m56.3 kg1070.2 smkg1060.9 4 4    8.27: Denote the final speeds as BA vv and and the initial speed of puck A as , 0 v and omit the common mass. Then, the condition for conservation of momentum is  0.45cos0.30cos 0 BA vvv  .0.45sin0.30sin0  BA vv  The 0.45 angle simplifies the algebra, in that sin ,0.45cos0.45  and so the B v terms cancel when the equations are added, giving sm3.29 0.30sin0.30cos 0    v v A From the second equation, s.m7.20 2  A v B v b) Again neglecting the common mass, ,804.0 s)m0.40( s)m20.7(s)m3.29( )21( ))(21(K 2 22 2 0 22 1 2      v vv K BA so 19.6% of the original energy is dissipated. 8.28: a) From .,)( 21 2211 21212211 mm vmvm vvmmvmvmvmvm    Taking positive velocities to the right, sm.3.00 1 v and sm1.20 2 v , so sm1.60v . b) 2 )sm60.1)(kg0.250kg0.500( 2 1 K 22 )sm20.1)(kg0.250( 2 1 )sm3.00)(kg0.500( 2 1  J.1.47 8.29: For the truck, kg,6320M and s,m10V for the car, kg1050m and sm15v (the negative sign indicates a westbound direction). a) Conservation of momentum requires mvMVvmM    )( , or eastbound.sm4.6 kg)1050kg6320( )sm15)(kg1050()sm10)(kg6320(      v b) .sm5.2 kg6320 )sm15kg)(1050(      M mv V c) kJ281KE for part (a) and kJ138KE for part (b). s.m1.3 )kg195( )sm2.7)(kg85( sm0.5 )kg195( )sm8.8)(kg110(   y x v v 8.30: Take north to be the x-direction and east to be the y-direction (these choices are arbitrary). Then, the final momentum is the same as the intial momentum (for a sufficiently muddy field), and the velocity components are The magnitude of the velocity is then ,sm9.5)sm1.3()sm0.5( 22  at an angle or arctan    32 0.5 1.3 east of north. 8.31: Use conservation of the horizontal component of momentum to find the velocity of the combined object after the collision. Let +x be south. P x is constant gives J0300.0 J0020.0)s100.0)(kg400.0( J0320.0)s600.0)(kg150.0()s200.0)(kg250.0( north),scm10.0(scm0.10 )kg400.0()s600.0)(kg150.0()sm200.0)(kg250.0( 12 2 2 1 2 2 2 1 2 2 1 1 22 2      KKK K K vv v x x Kinetic energy is converted to thermal energy due to work done by nonconservative forces during the collision. 8.32: (a) Momentum conservation tells us that both cars have the same change in momentum, but the smaller car has a greater velocity change because it has a smaller mass. (b) The occupants of the small car experience 2.5 times the velocity change of those in the large car, so they also experience 2.5 times the acceleration. Therefore they feel 2.5 times the force, which causes whiplash and other serious injuries. car)(largeV2.5 kg1200 kg3000 car)(largecar)(small    V V m M v vmVM 8.33: Take east to be the x-direction and north to be the y-direction (again, these choices are arbitrary). The components of the common velocity after the collision are h.km33.33 kg)4200( )hkm0.50(kg)2800( hkm67.11 kg)4200( )hkm 35.0(kg)1400( y       v v x The velocity has magnitude hkm3.35h)km33.33(h)km67.11( 22  and is at a direction arctan      7.70 67.11 33.33 south of west. . the square of the change of the speed. 8.11: a) With ,0 1 t ,)sN1000.2()sN 1080 .0( 3 2 292 2 7 0 2 ttdtFJ t xx   which is 18.8 smkg  , and so the. m/s,kg15.41m/s)0.5030coss)m0.65((kg)145.0(   so the x- component of the average force is N, 108. 81 s1075.1 m/skg41.15 3 3     and the y-component of the force is